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Let $\Omega$ be an open bounded convex set in $\mathbb {R}^n$
Let $\rho$ be a density function on $\Omega$ with $\rho >0$ almost everywhere in $\Omega$.

The center of mass of $\Omega$ is given by:

$$c = \dfrac {\int_\Omega y \rho(y)dy}{\int_\Omega \rho(y)dy}$$

Is it true then that $c\in \bar \Omega $?

It seems intuively true, but how can one prove it?

user74261
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3 Answers3

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It is true that $c \in \Omega$, but it is a pain to prove (I have only been able to prove it by induction on dimension).

Here I just show that $c \in \bar{\Omega}$.

The key result (using the Hahn Banach theorem) is that if $x \notin \bar{\Omega}$, there is some linear functional $\phi$ and a constant $\beta$ such that $\phi(x) <\beta$ and $\phi(y) \ge \beta$ for all $y \in \bar{\Omega}$.

So, suppose $c \notin \bar{\Omega}$, then let $\phi$ be the functional above such that $\phi(c) <\beta$.

Then $\phi(c) = \phi({ \int_\Omega y \rho(y) dy \over \int_\Omega \rho(y) dy}) = { \int_\Omega \phi(y) \rho(y) dy \over \int_\Omega \rho(y) dy} \ge \beta$, a contradiction. Hence $c \in \bar{\Omega}$.

copper.hat
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  • Is it also true that $c$ lies actually in $\Omega$, not on the boudary of $\Omega$?It seems very improbable that no matter what $\rho$ is, $c$ lies on the "edge" of $\Omega$ – user74261 May 01 '15 at 02:56
  • Since $\Omega$ is open, and $c \in \Omega$, then it does not lie in the boundary. The key here is that $\rho$ (divided by the total mass) is a probability measure. – copper.hat May 01 '15 at 02:57
  • It is true for any convex, bounded $\Omega$ that $c \in \Omega$, but it is easy to prove the $c \in \bar{\Omega}$ case, so that is all I did above. This is a result that I have often seen stated, but never proved. – copper.hat May 01 '15 at 02:59
  • I see. Pardon me if I sounded too aggressive :) BTW, why are you saying $\rho$ is a probablity measure? It does not integrate to 1 necessarily – user74261 May 01 '15 at 03:01
  • You weren't. It is a good question. – copper.hat May 01 '15 at 03:01
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    Why did you unaccept this answer after 5 years? – copper.hat Jun 17 '20 at 02:09
  • Why was this downvoted? – copper.hat Jun 19 '20 at 18:13
  • @copper.hat, Hello, may I know if being open is crucial here? Say, if $\Omega$ is bounded convex only, and define $c$ as above, then is it true that $c\in\overline{\Omega}$? – user284331 Jun 02 '22 at 02:07
  • No. I imagine one just needs the denominator to be non zero, this will be true, for example, if $\Omega$ contains an open set. – copper.hat Jun 02 '22 at 03:51
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I think I have another solution. We have that $\Omega$ is bounded, convex, and open.

Up to translation, we might assume $0 \in \Omega$. Consider the Minkowski functional of $\Omega$: $$N(x) = \inf\left\lbrace t >0 : \frac x t \in \Omega\right\rbrace.$$ The Minkowski functional $N$ is convex because $$N(x+y)\leq N(x)+N(y)\quad \text{and}\quad N(\alpha x) = \alpha N(x)$$ for $x,y \in \mathbb R^n$, $\alpha \geq 0$.

If $|\cdot|$ is the usual $\mathbb R^n$ norm, we have that $$|x|/R \leq N(x) \leq |x|/r,$$ where $$r = \inf\{|\omega|: \omega \in \Omega\},$$ $$R = \sup\{|\omega|: \omega \in \Omega\}.$$

Therefore, $N$ is continuous.

Note that $$\Omega:=\{ x \in \mathbb R^n: N(x)<1\}.$$

Applying Jensen's inequality, we obtain

$$N(c) \leq \frac{\int_{\Omega} N(x) \rho(x) dx}{\int_{\Omega} \rho(x) dx}< \frac{\int_{\Omega} \rho(x) dx}{\int_{\Omega} \int \rho(x) dx} = 1.$$

Thus, $c \in \Omega.$

Hugo
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If a half plane $H=\{x; \langle a,x\rangle -b\geq 0\}$ contains $C$ it contains $c=\int_C xdx/\int_C dx.$ To see this observe that $$\langle a,c\rangle -b=\frac{1}{\int_Cdx}\int_C (\langle a,x\rangle -b)dx\geq 0.$$ Finally, a closed convex set $C$ is the intersection of all the closed half planes $H$ such that $H\supset C.$ Hence $c\in C.$

Gérard Letac
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